I want to distinguish between numbers with/without a dot attached:

>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> re.compile(r'ch \d{1,}[.]').findall(text)
['ch 1.', 'ch 23.']
>>> re.compile(r'ch \d{1,}[^.]').findall(text)
['ch 23', 'ch 4 ', 'ch 56 ']

I can guess why the 'ch 23' appears in the second list. But how to get rid of it?

--Jach

Jach Feng <jfong@ms4.hinet.net> wrote:
> I want to distinguish between numbers with/without a dot attached:
>
>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>> re.compile(r'ch \d{1,}[.]').findall(text)
> ['ch 1.', 'ch 23.']
>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> ['ch 23', 'ch 4 ', 'ch 56 ']
>
> I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
>
> --Jach

Does

>>> re.findall(r'ch\s+\d+(?![.\d])',text)

do what you want? This matches "ch", then any nonzero number of
whitespaces, then any nonzero number of digits, provided this is not
followed by a dot or another digit.

Neil 在 2021年8月5日 星期四下午6:36:58 [UTC+8] 的信中寫道：
> Jach Feng <jf...@ms4.hinet.net> wrote:
> > I want to distinguish between numbers with/without a dot attached:
> >
> >>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>>> re.compile(r'ch \d{1,}[.]').findall(text)
> > ['ch 1.', 'ch 23.']
> >>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> > ['ch 23', 'ch 4 ', 'ch 56 ']
> >
> > I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
> >
> > --Jach
> Does
>
> >>> re.findall(r'ch\s+\d+(?![.\d])',text)
>
> do what you want? This matches "ch", then any nonzero number of
> whitespaces, then any nonzero number of digits, provided this is not
> followed by a dot or another digit.

Yes, the result is what I want. Thank you, Neil!

The solution is more complex than I expect. Have to digest it:-)

--Jach

On Thu, 5 Aug 2021 02:40:30 -0700 (PDT), Jach Feng <jfong@ms4.hinet.net> wrote:
I want to distinguish between numbers with/without a dot attached:
>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> re.compile(r'ch \d{1,}[.]').findall(text)
['ch 1.', 'ch 23.']
>>> re.compile(r'ch \d{1,}[^.]').findall(text)
['ch 23', 'ch 4 ', 'ch 56 ']
I can guess why the 'ch 23' appears in the second list. But how to get
rid of it?

>>> re.findall(r'ch \d+[^.0-9]', "ch 1. is ch 23. is ch 4 is ch 56 is ")
['ch 4 ', 'ch 56 ']

--
To email me, substitute nowhere->runbox, invalid->com.

Le 05/08/2021 à 11:40, Jach Feng a écrit :
> I want to distinguish between numbers with/without a dot attached:
>
>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>> re.compile(r'ch \d{1,}[.]').findall(text)
> ['ch 1.', 'ch 23.']
>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> ['ch 23', 'ch 4 ', 'ch 56 ']
>
> I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
>
> --Jach
>

>>> import re

>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'

>>> re.findall(r'ch \d+\.', text)
['ch 1.', 'ch 23.']

>>> re.findall(r'ch \d+(?!\.)', text) # (?!\.) for negated look ahead
['ch 2', 'ch 4', 'ch 56']

Le 05/08/2021 à 17:11, ast a écrit :
> Le 05/08/2021 à 11:40, Jach Feng a écrit :
>> I want to distinguish between numbers with/without a dot attached:
>>
>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>> ['ch 1.', 'ch 23.']
>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>
>> I can guess why the 'ch 23' appears in the second list. But how to get
>> rid of it?
>>
>> --Jach
>>
>
> >>> import re
>
> >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>
> >>> re.findall(r'ch \d+\.', text)
> ['ch 1.', 'ch 23.']
>
> >>> re.findall(r'ch \d+(?!\.)', text)   # (?!\.) for negated look ahead
> ['ch 2', 'ch 4', 'ch 56']

ops ch2 is found. Wrong

Le 05/08/2021 à 17:11, ast a écrit :
> Le 05/08/2021 à 11:40, Jach Feng a écrit :
>> I want to distinguish between numbers with/without a dot attached:
>>
>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>> ['ch 1.', 'ch 23.']
>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>
>> I can guess why the 'ch 23' appears in the second list. But how to get
>> rid of it?
>>
>> --Jach
>>
>
> >>> import re
>
> >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>
> >>> re.findall(r'ch \d+\.', text)
> ['ch 1.', 'ch 23.']
>
> >>> re.findall(r'ch \d+(?!\.)', text)   # (?!\.) for negated look ahead
> ['ch 2', 'ch 4', 'ch 56']

import regex

# regex is more powerful that re

>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'

>>> regex.findall(r'ch \d++(?!\.)', text)

['ch 4', 'ch 56']

## ++ means "possessive", no backtrack is allowed

ast 在 2021年8月5日 星期四下午11:29:15 [UTC+8] 的信中寫道：
> Le 05/08/2021 à 17:11, ast a écrit :
> > Le 05/08/2021 à 11:40, Jach Feng a écrit :
> >> I want to distinguish between numbers with/without a dot attached:
> >>
> >>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>>>> re.compile(r'ch \d{1,}[.]').findall(text)
> >> ['ch 1.', 'ch 23.']
> >>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> >> ['ch 23', 'ch 4 ', 'ch 56 ']
> >>
> >> I can guess why the 'ch 23' appears in the second list. But how to get
> >> rid of it?
> >>
> >> --Jach
> >>
> >
> > >>> import re
> >
> > >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >
> > >>> re.findall(r'ch \d+\.', text)
> > ['ch 1.', 'ch 23.']
> >
> > >>> re.findall(r'ch \d+(?!\.)', text) # (?!\.) for negated look ahead
> > ['ch 2', 'ch 4', 'ch 56']
> import regex
>
> # regex is more powerful that re
> >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>> regex.findall(r'ch \d++(?!\.)', text)
>
> ['ch 4', 'ch 56']
>
> ## ++ means "possessive", no backtrack is allowed
Can someone explain how the difference appear? I just can't figure it out:-(

>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> re.compile(r'ch \d+[^.]').findall(text)
['ch 23', 'ch 4 ', 'ch 56 ']
>>> re.compile(r'ch \d+[^.0-9]').findall(text)
['ch 4 ', 'ch 56 ']

--Jach

Il 05/08/2021 11:40, Jach Feng ha scritto:
> I want to distinguish between numbers with/without a dot attached:
>
>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>> re.compile(r'ch \d{1,}[.]').findall(text)
> ['ch 1.', 'ch 23.']
>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> ['ch 23', 'ch 4 ', 'ch 56 ']
>
> I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
>
> --Jach
>

import re

t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)

res = r.findall(t)

dot = [x[1] for x in res if x[1] != '']
udot = [x[0] for x in res if x[0] != '']

print(f"dot: {dot}")
print(f"undot: {udot}")

out:

dot: ['ch 4', 'ch 56']
undot: ['ch 1.', 'ch 23.']

Le 06/08/2021 à 02:57, Jach Feng a écrit :
> ast 在 2021年8月5日 星期四下午11:29:15 [UTC+8] 的信中寫道：
>> Le 05/08/2021 à 17:11, ast a écrit :
>>> Le 05/08/2021 à 11:40, Jach Feng a écrit :

>> import regex
>>
>> # regex is more powerful that re
>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>> regex.findall(r'ch \d++(?!\.)', text)
>>
>> ['ch 4', 'ch 56']
>>
>> ## ++ means "possessive", no backtrack is allowed

> Can someone explain how the difference appear? I just can't figure it out:-(
>

+, *, ? are greedy, means they try to catch as many characters
as possible. But if the whole match doesn't work, they release
some characters once at a time and try the whole match again.
That's backtrack.
With ++, backtrack is not allowed. This works with module regex
and it is not implemented in module re

with string = "ch 23." and pattern = r"ch \d+\."

At first trial \d+ catch 23
but whole match will fail because next character is . and . is not
allowed (\.)

A backtrack happens:

\d+ catch only 2
and the whole match is successful because the next char 3 is not .
But this is not what we want.

with ++, no backtrack, so no match
"ch 23." is rejected
this is what we wanted

Using re only, the best way is probably

re.findall(r"ch \d+(?![.0-9])", text)
['ch 4', 'ch 56']

jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
> Il 05/08/2021 11:40, Jach Feng ha scritto:
> > I want to distinguish between numbers with/without a dot attached:
> >
> >>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>>> re.compile(r'ch \d{1,}[.]').findall(text)
> > ['ch 1.', 'ch 23.']
> >>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> > ['ch 23', 'ch 4 ', 'ch 56 ']
> >
> > I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
> >
> > --Jach
> >
> import re
> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>
> res = r.findall(t)
>
> dot = [x[1] for x in res if x[1] != '']
> udot = [x[0] for x in res if x[0] != '']
>
> print(f"dot: {dot}")
> print(f"undot: {udot}")
>
> out:
>
> dot: ['ch 4', 'ch 56']
> undot: ['ch 1.', 'ch 23.']
> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
That's an interest solution! Where the '|' operator in re.compile() was documented?

--Jach

Il 06/08/2021 12:57, Jach Feng ha scritto:
> jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
>> Il 05/08/2021 11:40, Jach Feng ha scritto:
>>> I want to distinguish between numbers with/without a dot attached:
>>>
>>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>>> ['ch 1.', 'ch 23.']
>>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>>
>>> I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
>>>
>>> --Jach
>>>
>> import re
>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>>
>> res = r.findall(t)
>>
>> dot = [x[1] for x in res if x[1] != '']
>> udot = [x[0] for x in res if x[0] != '']
>>
>> print(f"dot: {dot}")
>> print(f"undot: {udot}")
>>
>> out:
>>
>> dot: ['ch 4', 'ch 56']
>> undot: ['ch 1.', 'ch 23.']
>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
> That's an interest solution! Where the '|' operator in re.compile() was documented?
>
> --Jach
>

I honestly can't tell you, I've been using it for over 30 years. In any
case you can find some traces of it in the "regular expressions quick
reference" on the site https://regex101.com (bottom right side).

Il 06/08/2021 16:17, jak ha scritto:
> Il 06/08/2021 12:57, Jach Feng ha scritto:
>> jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
>>> Il 05/08/2021 11:40, Jach Feng ha scritto:
>>>> I want to distinguish between numbers with/without a dot attached:
>>>>
>>>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>>>> ['ch 1.', 'ch 23.']
>>>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>>>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>>>
>>>> I can guess why the 'ch 23' appears in the second list. But how to
>>>> get rid of it?
>>>>
>>>> --Jach
>>>>
>>> import re
>>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>>>
>>> res = r.findall(t)
>>>
>>> dot = [x[1] for x in res if x[1] != '']
>>> udot = [x[0] for x in res if x[0] != '']
>>>
>>> print(f"dot: {dot}")
>>> print(f"undot: {udot}")
>>>
>>> out:
>>>
>>> dot: ['ch 4', 'ch 56']
>>> undot: ['ch 1.', 'ch 23.']
>>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>> That's an interest solution! Where the '|' operator in re.compile()
>> was documented?
>>
>> --Jach
>>
>
> I honestly can't tell you, I've been using it for over 30 years. In any
> case you can find some traces of it in the "regular expressions quick
> reference" on the site https://regex101.com (bottom right side).
>
....if I'm not mistaken, the '|' it is part of normal regular
expressions, so it is not a specific extension of the python libraries.
Perhaps this is why you don't find any documentation on it.

On 2021-08-06, jak <nospam@please.ty> wrote:
> Il 06/08/2021 16:17, jak ha scritto:
>> Il 06/08/2021 12:57, Jach Feng ha scritto:
>>> That's an interest solution! Where the '|' operator in re.compile()
>>> was documented?
>>
>> I honestly can't tell you, I've been using it for over 30 years. In any
>> case you can find some traces of it in the "regular expressions quick
>> reference" on the site https://regex101.com (bottom right side).
>
> ...if I'm not mistaken, the '|' it is part of normal regular
> expressions, so it is not a specific extension of the python libraries.
> Perhaps this is why you don't find any documentation on it.

The Python documentation fully describes the regular expression syntax
that the 're' module supports, including features that are widely
supported by different regular expression systems and also Python
extensions. '|' is documented here:
https://docs.python.org/3/library/re.html#index-13

jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
> Il 05/08/2021 11:40, Jach Feng ha scritto:
> > I want to distinguish between numbers with/without a dot attached:
> >
> >>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>>> re.compile(r'ch \d{1,}[.]').findall(text)
> > ['ch 1.', 'ch 23.']
> >>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> > ['ch 23', 'ch 4 ', 'ch 56 ']
> >
> > I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
> >
> > --Jach
> >
> import re
> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>
> res = r.findall(t)
>
> dot = [x[1] for x in res if x[1] != '']
> udot = [x[0] for x in res if x[0] != '']
>
> print(f"dot: {dot}")
> print(f"undot: {udot}")
>
> out:
>
> dot: ['ch 4', 'ch 56']
> undot: ['ch 1.', 'ch 23.']
The result can be influenced by the order of re patterns?

>>> import re
>>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M).findall(t)
[('ch 1.', ''), ('ch 23.', ''), ('', 'ch 4'), ('', 'ch 56')]

>>> re.compile(r'(ch +\d+)|(ch +\d+\.)', re.M).findall(t)
[('ch 1', ''), ('ch 23', ''), ('ch 4', ''), ('ch 56', '')]

--Jach

Il 07/08/2021 04:23, Jach Feng ha scritto:
> jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
>> Il 05/08/2021 11:40, Jach Feng ha scritto:
>>> I want to distinguish between numbers with/without a dot attached:
>>>
>>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>>> ['ch 1.', 'ch 23.']
>>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>>
>>> I can guess why the 'ch 23' appears in the second list. But how to get rid of it?
>>>
>>> --Jach
>>>
>> import re
>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>>
>> res = r.findall(t)
>>
>> dot = [x[1] for x in res if x[1] != '']
>> udot = [x[0] for x in res if x[0] != '']
>>
>> print(f"dot: {dot}")
>> print(f"undot: {udot}")
>>
>> out:
>>
>> dot: ['ch 4', 'ch 56']
>> undot: ['ch 1.', 'ch 23.']
> The result can be influenced by the order of re patterns?
>
>>>> import re
>>>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>> re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M).findall(t)
> [('ch 1.', ''), ('ch 23.', ''), ('', 'ch 4'), ('', 'ch 56')]
>
>>>> re.compile(r'(ch +\d+)|(ch +\d+\.)', re.M).findall(t)
> [('ch 1', ''), ('ch 23', ''), ('ch 4', ''), ('ch 56', '')]
>
> --Jach
>
Yes, when the patterns intersect each other as in your case. the
difference between the 2 patterns is the "." in addition. The logical or
does not continue checking when the condition is satisfied, so it is a
good idea, in these cases, to search for the most complete patterns
before the others.

Il 07/08/2021 11:18, jak ha scritto:
> Il 07/08/2021 04:23, Jach Feng ha scritto:
>> jak 在 2021年8月6日 星期五下午4:10:05 [UTC+8] 的信中寫道：
>>> Il 05/08/2021 11:40, Jach Feng ha scritto:
>>>> I want to distinguish between numbers with/without a dot attached:
>>>>
>>>>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>>>> re.compile(r'ch \d{1,}[.]').findall(text)
>>>> ['ch 1.', 'ch 23.']
>>>>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
>>>> ['ch 23', 'ch 4 ', 'ch 56 ']
>>>>
>>>> I can guess why the 'ch 23' appears in the second list. But how to
>>>> get rid of it?
>>>>
>>>> --Jach
>>>>
>>> import re
>>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> r = re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M)
>>>
>>> res = r.findall(t)
>>>
>>> dot = [x[1] for x in res if x[1] != '']
>>> udot = [x[0] for x in res if x[0] != '']
>>>
>>> print(f"dot: {dot}")
>>> print(f"undot: {udot}")
>>>
>>> out:
>>>
>>> dot: ['ch 4', 'ch 56']
>>> undot: ['ch 1.', 'ch 23.']
>> The result can be influenced by the order of re patterns?
>>
>>>>> import re
>>>>> t = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>>>> re.compile(r'(ch +\d+\.)|(ch +\d+)', re.M).findall(t)
>> [('ch 1.', ''), ('ch 23.', ''), ('', 'ch 4'), ('', 'ch 56')]
>>
>>>>> re.compile(r'(ch +\d+)|(ch +\d+\.)', re.M).findall(t)
>> [('ch 1', ''), ('ch 23', ''), ('ch 4', ''), ('ch 56', '')]
>>
>> --Jach
>>
> Yes, when the patterns intersect each other as in your case. the
> difference between the 2 patterns is the "." in addition. The logical or
> does not continue checking when the condition is satisfied, so it is a
> good idea, in these cases, to search for the most complete patterns
> before the others.
>
>
PS
.... the behavior of the logical or that I have described is not typical
of regular expressions but it is common in all programming languages.